How to show #f(x)=|x|# It is differentiable everywhere except at the point x = 0 ?

1 Answer
MattyMatty
Feb 9, 2018

#"See explanation"#

Explanation:

#"Apply the definition of |x| : "#
#f(x) = |x| => #
#{ ( f(x) = x, x >= 0 ), ( f(x) = -x, x<= 0 ) :}#
#"Now derive : "#
#{ ( f '(x) = 1, x >= 0), ( f '(x) = -1, x <= 0 ) :}#
#"So we see there is a discontinuity in x=0 for f '(x)."#
#"For the rest, it is differentiable everywhere."#

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